Ganguli, Theory of Plane Curves, Chapter 7

8/14/2019 Ganguli, Theory of Plane Curves, Chapter 7

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CHAPTER VIICHAHACTERl8TlCS OF C[RVES

146. PL~CKER'S EQL\TlOXS:"'IVeshall denote the degree of a curve

" class "" number of nodes" " "cusps

" "double tangents" " " stationary "

and the deficiency by p.The six quantities n, 'In, 0, K, r, t are called Plucker's

numbers or the Characterietics of the curve.Then we have m=n(n-l)-2o-'~K. (121)

t=3n(n-2)-60-SK. (112)

by n" 111., 0" K" r"

(1)(2)

The corresponding numbers for the reciprocal curve areobtains d by interchanging n and 'In, rand 0, and K.

Thus, from the reciprocal curve, we obtain-n=m(m-l)-2r-3 (3)

From equations (1), (2), (3) and (4), we may expresso in terms of 'In, r, and r in terms of n, 0, K:-Thus, 2r=n(n-2)(n2 -9)-2(n2 -n-6)(20+3K)

+40(0-1) +120K+9K(K-l)20=m(1Jt- 2) ('In2 - 9) - 2( 'In2 -?n- 6) (2r+ 3)

+4r(r-l) +12n+9t(t-l)

(5)

(6)


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182 THEORY OP PLANE CURVES'1'he six equations 1-6 are called Plucker' Equations,"It is readily seen that a curve has the same characteristics

as any projection of the curve, but curves with the samecharacteristics are not necessarily the projections oeachother. All curves with the same characteristics, however,are said to be of the same type.

147. The above formula (5) IS cumbrous in orm andIS not geometrically intelligible. 'INe give here the follow-ing simpler form to the equation (5), which can be geo-metrically interpreted.

It can easily be shown, that two double tangents coi n-cido with each of the tangents drawn rom a node to acurve, three coincide with each tangent drawn 1'0m acusp, [owr coincide with each line joining two nodes, sicoincide with each line joining a node to a cusp, and ninecoincide with each line joining two cusps.

Now 01' a non-singular curve, the number of doubletangents is given by T=in(n-2)(n' -9) and if the curvehas 0 nodes and K cusps, this number will be reduced.Now, (m-4) tangents can be drawn from each node to thecurve, and (1n-3) tangents from each cusp.

:. 20(m-4) doubletangents coincide withthe tangents drawnfrom the nodes, and3K(m-3) tangents coin-cide with those drawnfrom the cusps. Thenumber of lines joiningthe nodes is~-0(0-1),

4

* Pliicker-Solution


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VH,\\l.ACTERISTlC:; 01' CURVES 183and consequently theyare equivalent to2il(8-1)double tangents. TheDumber of lines joiningthe nodes with thecusps is ilK , and theyare equivalent to 60Kdouble tangents. Final-ly, there are tK(K-l)lines joining tqe cusps, 3and they are equivalentto tK(K-l) doubletangents. Thus thenumber of double tangents to a curvewith Snodes and K cusps is-

of order n, class m,

T=tn(n-2(n' -9)-2il(m-4) -3K(m-3)-2S(S-I)

This equation is, in fact, equivalent to (5). A similarexpression may be obtained for the equation (6).

148. THE BITA}!GENTIAL CURVE:We can however directly determine the nnmber of

bitangents of a non-singular n-ic, and for this purpose, wefind a curve 1\=0 which intersects the n-ic in the pointsof contact of its bitangents. The curve 1\ =0 is called thebitangential curve, and its order for a non-singular curveis, in general, (n-2)(n2 -9).

* Prof. Cayley first determined the curve passing through the pointsof contact of bitangents-Crelie, Ed. 34 (1847), p. 37. Anothermethod for determining this curve has been given bySalmon-QuarterlyJournal of Mathematics, Vol. III, p. 317, and demonstrated by Oayley-Phil. Transactions (1859), p, 193,and (1861), p. 357.


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184 THEORY OF PLANE CURVESThe roots in 'A .ip . of the equation (1) of 63 give the

points where the line joining the points ( ...,',y', Z') and( .1 ;" , y" , z") meet the curve F=O. We have seen in that articlethat if Lv',y', z') lies on the curve andfe", y"~ /') is a point onthe tangent, F(.rJ, y', z')=F'=O and ,t:.F'=O. If thetangent at (,,', y', z') touches the curve elsewhere, then,making F' =0 and ,t:.F' =0 in that. equation, the red ueedequation of order (n-2) must have equal roots. Conse-quently, the discriminant D of this reduced equationmust vanish for (.,', y ', z') and (.:", y" , z"), But, as inthe case of a point of inflexion ,t:.F'=O and ,t:.'~"=O,and also ,t:.SF' contains ,t:.F' as a factor, in the case ofa bitangent the discriminant D must contain ,t:.F'=Oas a factor, and the condition thus obtained is thecondition that the point (e', y ', z') shall be n.point of contactof a bitangent.

Now, the reduced equation takes the form-

The discriminant D of this contains terms of the form(,t:.tF'),-sF'-s, and therefore, D is of order (n+2)(n-3)in (:t", y" , z"), of order (n-2)(n-3) in (/, y ', . ') , and oforder 2(n-3) in the co-efficients of the original equation.

But all the intersectionsofD=Oand ,t:.F'=O will coincidewith (x', y', z'). For, the equation of the tangents drawn from( : 1 : ' , y ', . ') ( 67) is of the form k,t:.F'+D(,t:.'F')" =0.Hence, these tangents are intersected by ,t:.F' =0 in noother point than (x', y', z'). Thus, if we put ,t:.F'=O inth is equation. we see that ,t:.F' can neither meet D nor ,t:.2F'in any other point than (.,i, y ', z ') .

Now, therefore, we have two curves ,t:.F'=O and D=Oof orders 1, (n+2)(n-3) respectively ir; Ce" , y" , z" ) aad oforders (n-I) and (n-2)(1-3) in (J", y', z'), and the


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CIURACTERISTICS 01" CUll VEl; 185coincide with (;e ', y', z'). Then, by a known lemma." thecondition that the cur-ves have other common points is oforder

(n-2)(n-:3) + (n-l)(11 +2)(n-3)- (n+ 2)(n-3)i.e., of order (n-2)(n' -9) in C v ' , y', Z l ) .This condition 1\=0 is, therefore, of order (n+2)(n-3) inthe co-efficients of .6 . F', of the first order in the co-efficientsof D, and consequently, of order (n+4)(n-3) in the co-efficients of the original equation.

Hence, the points of contact C " ' , y ', c ') of the bitangentsof the curve F==O are the points where 1\ =0 meets it,and their number is therefore n(n-2)(ni-9).

But there are two of these points on each bitangent,the number of bitangents is, therefore, ~n(n-2) (n' -9).

Salmon has given an expression of the bitangentia.l curve1\ =0 for a general curve of order n.t149. From Plucker's formulre various other important

results can be deduced:From (3) and (4), by eliminating T we obtain

K-3n=t-3m, or, K-t=3(n-m) (7)The same equation also follows from (1) and (2) by

eliminating 8 . " r e see, therefore, that the four equationsare not independent, but they are equivalent to threeequations only.

From (1) and (3) it follows by subtraction thatnO -?n' =2(8-T) +3(K-t)

or, n2 -m'=2(8-T) +9 :n-m).:. (n-m)(n+m-9)=2(8-T) (8)

* Salmon-H. P. Curves, 381.t Salmon-H. P. Curves, 38!-92.


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186 THEORY OF PT,ANE OURVESFrom (5) and (8), we obtain-s-

Un' -rn' +3(n-m,)]=U2(S-T) +4(K-t)]=S-T+2(K-t)

~n(n+3)-S-2K=trn(m+3)-T-2t (9)r,150. The equation (9) has a very simple geometrical

interpretation :-We have seen ( 21) that a curve is uniquely determined

oy tn(n+3) given points, or, in other words, the equationof a curve of the nth degree can be made to satisfy tn(n+3)conditions.

But the existence of a node reduces the conditions byone, and that of a cusp by two. Therefore, the number ofpoints determining a curve of the nth degree, with S nodesand Kcusps, is tn:n+3)-S-2K, and the above equationsays that this number is equal to t1n(m+3)-T-2t, which ISthe equivalent expression for the reciprocal curve.

Hence, a c'urve and its reciprocal polar are determined bythe same member of conditions, as is otherwise evident, sincewhen a curve is given, its reciprocal is determined. (See 62).

Again, we have-

H n i _rni -3(n-rn)} =Hn-l)(n-2)-Hrn-l)(rn-2).The left-hand side, by (7) and (8), is equivalent to

H2(S-T)+2( K-t)]=S-T+K-L:. t(n-l)(n-2)-S-K-=:Hm-l)(m-2)-T-L=P (10)

The number p if!called the Deficiency of the curve.The eqnation (10) says that a curve and its reciprocal


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Cl:iAltACT~RlSTics OF CUltVES I S i

we can write the formulm in the followiug simple forms :-

Im+K-2nI n+t-2m

2(p-l)= iI n(n-3)-2(8+K)lm(m-3)-2(T+t)

Prof. Cayley * puts t+8n = K+3m = aThen all the Plucker's numbers can be expressed interms of three only, namely, n, m, a.Thus,

28=n2-n+8m-3a,

151. THE POI~T AX)) LnE DEnCIE};CII(S:We have seen that a curve does not, in general, possess

singular points, unless certain conditions are satisfied bythe constants in the equation. But the general equationrepresents a curve which ordinarily possesses certain doubleor stationary tangents. Thus, double tangents andstationary tangents may be reckoned as the ordinary singu-larities of a curve whose point-equation is given, while allother higher multiple tangents may be regarded asextraordinary singularities, the presence of which requirescertain conditions to be fulfilled by the constants ill theequation. But, if the tangential equation of a curve is given,the curve ordinarily possesses double and stationary pointsaud CUISP;;, but no singular tangents. Hence, double andstationary poiuts are ordinary singularities of curves givenby its tangential equation, but the presence of highersingular points are subject to certain conditions. 'l'herefore,these oi-dinary sillguial'ities are such, that if auy curve

" Cuyley-Llual't"rly Jouruul, Vol XI, p _ 1~5.


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188 TI(EORY OF PLANE C U R V : I ! J Spossesses the one, its reciprocal will possess the reciprocalsingularity.

From all these considerations we are led to concludethat, if a curve has its maximum number of double pointsfor a curve of that order, it has also the maximum numberof double lines for a curve of that class. But it does notmean that the presence of double points on one leads to thepresence of double lines on the other. The presence ofmaximum number of double points on a curve reduces itsclass to suchan extent that the possible number of doubletangents is thereby diminished, and made the same as theactual number. Similarly, in a curve of given class, theexistence of the maximum number of double tangentsreduces the order to such an extent that the possible numberof double points is made the same as the actual number.Thus it is seen that for a curve of given order and class,the point deficiency and line deficiency are the same.

152. CURVES WITH THE SAME DEFICIENCY:We shall now prove the following general theorem:-If two curves have a one-to-one correspondence, i.e., are so

related that to any point of one corresponds a single point 01'tangent of the other, they have the same deficiency. *Let Sand S' be any two curves of orders nand n'

respectively, whose classes are rn and m', Let 8 and 8' bethe number of nodes and K, K' the number of cusps onthem. Let p and p' be their deficiencies respectively.

Let A and A' be any two fixed points in the plane, andP and P' two corresponding' points on S, S' respectively.Let C be the locus of the intersection of the lines AP andNP'.

The degree of C may be determined as follows :- This proof was simultaneously given hy Zen then (Com pt. rend

Ac. Se, Paris, Vol. 52,1869,1',742) and Bertini (Giorn. di Mat., Vol. 7,18ti9,pp. 105-l0ti).


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CIlAR,\CTEltISTlCS


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1 9 0 TlU;olty O~' PLAN!': ouuvss

on the locus C, It node 0[' a pail' of distinct points j butin neither case is AP an ordinary tangent to C.

In case (:3), according as the cusp on S corresponds toa cusp or a pair of coincident points on S', AP passesthrough a CURp on the locus C, or else is an ordinary tangent.

Thus ordinary tangents are obtained only in cases (1)and (3).

Now the class of S being m, the m tangents to Swillbe the tangents to C drawn from A .

Again, if there are l corresponding cusps on Sand S',the number of tangents to C in case (3) will be K-l. It isto be noted that there are l corresponding cusps on C.Hence, the total number of tangents, which can be drawnfrom A to the locus C, is 2n'+m+K-l.i.e., the class of C is 2n'+m+K-l ..Similarly, considering the number of tangents which

can be drawn to C from A', we find tor its class2n+m'+K'-l.

2n'+m+K-l = 2n+m'+K'-l.or,But m=n(n-l)-2o-3K and m'=n'(n'-1)-2o'-3K'.:. n(n-l)-2o-2K-~n = n'(n'-1)-2o'-2K' -2n'

i.e., v=vCor.: The deficiency of a curve and its reciprocal polt.'

is the same.

Ex. 1. The deficiency of a curve and its evolute is the same.Ex. 2. The Hessian, the Steinerian and the Cayleyan of a Cllrvehave the same deficiency.

Ex. 3. A curve has the same deticiency as its inverse and pedal


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CH.\ R.ICTERJ"TICS O f" CURY~;S 191153. From what has been said in 4G about multiple

points on a curve, it follows that if the curve has multiplepoints, the equations 1-6 still hold, subject to certainequivalent conditions. In fact, Plucker's equations are stillsatisfied, if a multiple point of order k be regarded asequivalent to t7.-(k-l) double points, and reciprocally, ak-ple tangent be replaced by t7.-(k-l) bi-tangents.

'rhus, if the curve has multiple points of orders" 1 ' k 2 ' " we havem=n(n-l) -28-8K-lk(k-l)

and t=8n(n-2) - 68-8K-3lk(k-l).Reciprocally,

n=m(m-l) -2T-3t-lk(lc-l)K=3m(m-2)-6T-8t-3lk(k-l)

where lextends over all the multiple points and tangents.It can easily be deduced that the deficiency of a curve

with only ordinary multiple points with distinct tangents isHm-2n+2).

For, in this case K=O,., p=Hn-l)(n-2)-!lk~k-l)

or, 2p=(n-I)(n-2)-lk(k-l)={n,n-l)-lk(k-l) }-2(n-l)

( 53)

=m-2n+2.p=Hm-211+2).

154. THE CHARACTERIS'l'TCS OF THE HESSfAN :


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192 1Hl>ORY o r PLANE CURVESpoints, the Hessian ordinarily has 0 .double points," andits Plueksrian characteristics are easily found to be-

n'=3,n-2), 8'=0, K'=O,m'=3(n-~)(3n-7)

T'='}-j-(n-l)( n-2)(n-3)(!:~n -8)an I L'=9(n-2)(3n-8) p'=H3n-7)(3n-8)where n', m', 8', K' etc., denote the Pliicker's numbers ofthe Hessian.

If, however, the original curve has nodes and cusps,each node is a node on the Hessian and each cusp is a triplepoint. Hence, these numbers must be modified accordingly,if the original curve has nodes and cusps.

155, THE CHARACTERISTICS OF TH8 STF.INERIAK : tThere is a (1, 1) correspondence between the Steinerian

and the Hessian, Hence, the deficiencies of the two curvesmust be the same. We have already found ( 96) that theclass of the Steinerian of a non-singular n-ic is 3(n-l )(n-2)and its order is 3(n-2) 2.

A point will be a node or cusp on the Steiner-ian, if it isa point whose first polar has two nodes or two cusps. Thenumber of first polars having a pair of nodes ( 98) is-

f(n-2)(n-3)(3n2-9n-5)and the number having two cusps is 12(n-2)(n-3).

Hence, the characteristics of the Steinerian of a non-singular curve are-

n'=3(n-2)2 m'=3(n-l)(n-2),8'=i(n-2)(n-3)(3n' -9n-5)

K'=12(n-2)(n-3)T'=Hn-2)(n-3)(3n' -3n-8)

,'=3(n-2)(4n-9) p'=H3n-7)(3n-8).* Pezza-Napoli, Rend, Vol. 22 (1883).t Steiner=-" Allgemeine, etc., Orelle, Bd, 47, p. 4.


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CHARACTER.ISTICS O~' CURVES 193156. THE CHARACTERISTICS OF THE CAYLEYAN : *This curve has evidently a (1, 1) correspondence with

the Hessian and with the Steinerian, and has therefore thesame deficiency.

We have already determined the class of this curve( 101) which also touches the inflexional tangents of theoriginal curve. It has no inflexions, in general, and thuswe obtain the following characteristics of the Cayleyan :

n'=3(n-2) (5n-ll) m'=3(n-I)(n-2)8'=--(n-2)(5n-I3)(5n'-I9n+I6)K'=I8(n-2}(2n-5) "=0

p'=t(3n-7)(3n-8)

157. THE CHARACTERISTICS OF THE. INVERSE CURVE:From what has been said with regard to the process of

inversion ( 15) and the properties of inverse curves( 139), it follows at once that a curve and its inversehave a one-to-one correspondence and consequently, thecharacteristics of the latter can be easily determined.

If (1)be the equation of a curve, that of its inverse t isuo(a:+y")"+k'u1(x+y,)n-l+ ... +k'u.=0 .. (2)Hence, the inverse curve has a multiple point of order n

at each of the circular points I and J, and has also ann-plepoint at the origin.

The degree of the inverse curve (2) is evidently 2n. Butif the origin is a k-ple point on (1), the degree of the inverse

it Olebsch-e-" Ueber eimge vou Steiner behandelte Curven-Crelle's Journal, Bd. 64, pp. 288-293.t A. S. Hart-Camb. and Dublin Matb. Journal, Vol. VII(l) (1863);


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194 THEORY OF PT.ANE CURVES(2) IS 2n-k. The degree will be further reduced, ifu., u._ u etc., contain some pOWArof r as a factor, i.e., if thecurve (1) has multiple points at I and .J .

Then, as before, n'=2nBut, if the origin IS a multiple point of order k , the

degree of the inverse will be 2n- k.A node on the given curve inverts into a node on the

inverse, and in addition, each of the points I and J and theorigin is an n-ple point on the inverse. Each of these threepoints is then equivalent to tn(n-l) nodes.

Hence, 3/=3+tn(n-l).Again, a cusp inverts into a cusp, so that we have K/=K_From these, the other characteristics of the inverse can

be easily calculated.Thus, m'=n'( n'-I) -23/ -3K'

=2n(2n-l)-23-3n(n-l)-3K={n(n- 1)-23-3K}+2n=m+2n.

t'=3n'(n/- 2) - 6 3/ - SK /=6n(2n -2)-63-9n(n-l)-8K=3n(n-2) +3n

Similarly, T'=2n(2n-7)+4mn+2Tand p'=p.

It is to be noted, however, that in these investigations,the curve is supposed to have only nodes and cusps and noother higher multiple points.

It will be shown later on that the foci of a curve invertinto the foci of the inverse curve, and that, if the origin is a


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CliAltAC1'ERISTlCS 0).' CURVES 1 9 5

focus on the curve, the circular points are cusps on theInverse.

If I and J are each a multiple point of order f, and theline at infinity a multiple tangent of order g , and, I, g ' denotereciprocal singularities, then the above results have to bemodified.

Thus, n'=2n-2j-g'm'=m+2n-2(2f+g')-(21 +g)

o'=Hn-2f)(n-2f-l)+(n-f-(l)(n-f-g'-I)+oand so on.

E, 1. A 2nic with >t.ple points at, I and J inverts into a curve ofl I same type.

E1l. 2. Show that a bitangent inverts into a circle having doublecontact with the inverse. Hence, find the uumber of circles passingthrough a given point and having double contact with a.curve.

E, 3. Prove that an inflexional tangent inverts into a circle ofcurvature of the inverse. Find the number of circles of curvature of agiven curve which pnss through a given point.E, 4. Shew that the curveC(Z2 + y2)' + 2(lz +my)(z' + y') + nz' + 2hzy + by' + 2gz + 2fy=O

inverts into a cubic through the poiuts I and J.ErII. 5. Prove that through auy point 0 on the above curve, three

real circles of curvature pass, besides the circle of curvature at 0, andthe three points of osculation lie or: a circle through O.

158. THE CHARACTERISTICS OF TH}J PEDAL:From what has been said in articles 137-39, it

follows that the fir-st positive pedal is the inverse of thepolar reciprocal curve. Hence, the characteristics of thepedal call be obtained from those of the polar reciprocalcurve by using the results of the preceding article, i.e.,


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196 THEORY OF PLANE CURVESin the results of that article, we have simply to interchangen and 'In, 0 and T, Land K. Thus,

n'=2m, m'=n+21n,,,'=t, T'=2?n(21n7) +4rnn+20,

p'=p.In cases of higher singularities, these numbers require

to be modified according to the nature of the singularity.If f, g, f', g', etc., have their significance as ill 157,

we have-n'=2rn-f-g, m'=n +2rn- 2(g +2j') -(2f+g')

and so on.Taylor defines the pedal t of a pair of curves as follows:The locus of the veriex of a right angle whose arms

envelope two curves of class 'In and class n respectivelymay be called the pedal of the two curves, or of the oneuur.t, the other, and the corresponding locus generated bythe vertex of any other constant angle may be called askew pedal. When one of the curves in the former casedegenerates into a point, we obtain the ordinary pedal ofthe curve.

Taylor says that the lines 01 and OJ may be regarded as perpendicular to everyone of the nt tangents of a given curve of class nt,which can be drawn from 1 and J respectively.

Each of I and J, therefore, is an m-ple point on the pedal and thishaving no other point at infinity is of order 2m. When 0 is at a focus,each of the lines 01 and OJ is a tangent and also perpendicular toitself. Hence, these lines making up the point circle at 0 belong to thelocus and the remaining factor is of order 2(n-l).

Taylor-Messenger of Math., Vol. 16 (1887), p. 4.t Dr. C. Taylor-Proc. of the Royal Soc. of London, Vol. 37 (1984),p.139.


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CliARAC1'ERISTICS OJ' CURVES 197Ell). 1. Pind the characteristics of the second, third, etc., pedals

of a given curve.Ell). 2. Find the characteristics of the first negative pedal.[This may be regarded as the polar reciprocal of tbe inverse curve

with respect to any point; and hence, the characteristics may beobtained from the results of 157.

n'=1n+2n m'=2n~'=211(2n-7)+4m1t+2T, etc.]

Ell). 3. Find the eharac terist.ics of the locus of the centre of a circlepassing through a given point and touching a given curve.

Ell). 4. Find the characteristics of the envelope of a circle whichpasses t.hrough a given point und whose centre moves on a curve.

159. THE CHAHACTEJUSTICS OF THE EVOWTE *:In order to determine the degree of the evolute, it issufficient if we examine the number of points in which the

line at infinity meets the evolute.Now, the points at infinity on the evolute arise (1) from

the points at infinity on the curve, (2) from the existenceof points of inflexion on the curve.

Corresponding to a point at infinity on the curve, wehave a cusp on the evolute, with the line at infinity as thecuspidal tangent.

Let M be any point on the line IJ, and M' its harmonicconjugate, then the normal at M is the line IJ ( 127).But if the consecutive points of the curve, antecedent andsubsequent to M be Land N, their normals are LM', ~M'.Hence, M' is a point through which three consecutivenormals, i.e., three consecutive tangents to the evolute pass,and is, therefore, a cusp with IJ for its tangent.

Now, the cuspidal tangent meets a curve in three conse-cutive points at a cusp, and the n points at infinity of the given

Steiner-" Dber algebraisehe Curven und Flltchen "-Crel1e,


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198 THEORY OF PLANE CURVEScurve give rise to the same number of cusps on the evolute,which are then met by the line at infinity in 3n points.

Again, a point at infinity on the evolute is the point ofintersection of two consecutive normals to the given curvewhich are parallel. The corresponding tangents to thegiven curve will therefore coincide, and the point of contactwill be an inflexion on the given curve. Therefore, thet points of inflexion on the given curve give rise to t pointsat infinity on the evolute.

Hence, the line at infinity intersects the evolute int+ 3n points, or in other words, the deqree of the euoluteis t+3n.

If the curve passes through either I or J, these give riseto no points at infinity on the evolute, and consequently, thedegree will be diminished by 3.

If, again, the line at infinity IJ touches the curve, thenormals for the two consecutive points in which it meetsthe curve coincide with IJ; and consequently two, consecu-tive tangents to the evolute coincide, i.e., there is a point ofinflexion on the evolute, having lJ for its tangent. But thistakes the place of two cusps which we have when IJ meetsthe curvs in distinct points, and the degree of the evolute 1Sreduced by three.

Hence, if each of the circular points is an f-ple point onthe curve and the line at infinity touches it at 9 points, thedegree of the evolute is-

n'=t+3n-3(2f+g)=a-3(2f+g)

160. The class of the evolute may be determined byconsidering the number of tangents which can be drawn toit from any point, or what is the same thing, by consideringthe number of normals which can be drawn from any pointto the given curve. We may examine the case when thepoint is at infinity.


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CHARACTERISTICS 01


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200 THEORY OP PLANJIJ CURVEScurve passes through I or J, the evolute has an inflexionaltangent passing through the same point.

Hence, in this case, t'=2f+gWe can now easily calculate the other characteristios of

the evolute by means of Plucker's formulae.Thus,Since,

K'=3n' +t' -3m'=3(,+3n)+t' -3(m+n).,'=0, K'=9n-3(m+n)+3,=3{2n-m+,}

When the curve touches the line at infinity at g points andhas f-ple point at each of Iand J, we have-

,,'=3n'+,'-3m'

And

whereFinally,

=3{,+3n-3(2f+g)} +(2f+g)-3(m+n-2f-g)=3a-3(m+n)-5(2f+g)

where a=t+3no'=-!{n" -n' +8m'-3a'}T'=i{m't -m' +81l'-3a'}

a'=t'+3n'=3a-8(2f+g)p'=!{n'+t'-2m'}+1 ( 150)=H,+3n-2(m+n)}+1=p

Ex. 1. Find the characteristics of the evolute of a'y =t'.n'=8, m'=8, p=O.

Ew . 2. Find the number of points on a curve where the osculatingcircle has a contact of the third order.

[The existence of a cusp of the evolute not lying on the line atinfinity indicates the coincidence of three consecutive tangents of theevolute, and consequently of the coincidence of three consecutivenormals to the curve, or in other words, corresponding to such a cnspon the evolute, we have a point on the curve where three consecutivenormals coincide, i.e., the osculating circle has a contact of the thirdorder. Thus the number of such points =('-n, since the n cusps ouIJ do not give any such point.

/('-n=5n-3m + 3,.]


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VIL\RACTEIIIS1'!VS (w CU\tV};,,: 20tE, 3. How many lines are normal to a curve at two points ~A bitangcnt on the ovolute corresponds to such a normal to thc curve.[Honco, thero are T' such points, But the 1/, normals corresponding

to the points at infinity on the curve coincide with the line at.infinity. Theroforo, excluding these .}n(n-l) normals, corresponilingto thl'l~PII points, whir-h coinr-irlr- with 1.1', thpl'I' P.IIHr,r, EL Cunvns :To determine the degree of the parallel cn rve, we pnt

1;=0 in the equation, which does nor affect the terms of thehighe"t degree in the equnt ion. TIl(' rosn lt of pntting 7.:=0is, however, the original curve written twice togethet- withthe two sets of 11 1 tangents rl ra w n from the circn lar point~ rand J to the curve,

Hence,Again, the number o f

parallel to ltny gi ven linedrawn to the Otiginal cur vc,

tangents which can he drawnis dou hie that. which can he so

To each inflexional tangent Oil th prig-inal c:oJ'J'f'sponcltwo on the parallel curve , and ther-efore ,'=2,.

From these we can easily calen latu the other c hnrne-teriatics.


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:W2 1'HIWRY O}


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CHAltACTE\tIS'l'ICS O~' CURVES

For, let "\ a.nd B he two points Ull the line IJ, harnioniceunjugates with respect to 1 and J. 'I'he two tangents fromA and B intersecting at Care theu perpendicular, andconsequently, C lies on the orthoptic locus. 1 then A andB approach I, U also approaches I, while C(lJ, AB) remainsharmonic.

If the tangents Iroiu A and B are not consecutive,(j becomes the point of contact of either tangent.

Now proceeding to the limit, we see that to each pail' uftangents to the given curve from I corresponds a branch ofthe orthoptic locus through I, and the tangent. to this branchis harmonic conjugate to 1.J with respect to these tangents.Since the class of the curve is In, the 1n tangents through Imay be taken to constitute tm('n-l) such pairs oftangents, and consequently, there i", the same number1-m(rn-l) of branches of the curve which pass through I ;or in other words, each of the points I and J is a multiplepoint of order -}1n(m-I). Further, it can be easily seenthat there is no other point of the locus on J J.

Hence, the degree of the orthoptic locus i8-n'=m(m-l).

16 :3 . '1 '0 find the class of the orthoptic locus, it issufficient to find the number of tangents which can bedrawn to the locus from J.

Since each of I and J IS ~L multiple point of order-}m(m-l), there are m(m-1) such tangents at J ( 65)which are to be regarded as tangents drawn from J to thecurve. In order to find the other tangents we proceed asfollows;

In the figure of 142, let PP' pass through J; then QRpasses through I, for PCQR,IJ) and P'(QR,IJ) areharmonic.

In the limit, Q, R become the poiuts of contact of theperpendicular tangents PQ, PR; and the tangent at P to


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the orthoptic locus pas~es through.1 while QR pa,sse,;through I,

If a line through 1 iut.ersects he gt ven curve at \l,and R, and tlte tangellt~ at Q and R meet in 0, let L be [Lpoint on IQR, such that O(QR,' Ill) iR harmonic. Nowconsider- the envelope of the line 01-


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CHAR.\c'fi'LtiSTiCS O~' CUR\')cS ~05It'=t' +:~n'=t1n+3C:-~1n-1nn-ln) +811~(1I!1)28'=".'2 -1'1' +811/' -Set' +:111')=n" -lOn' +t>m'-3t'=m' (in-I)' -10In(m-1)+Sm(m+n-:3)

-3m( t+ 3it-6).=m{(m+ 1)(m-2)' -2T 1=-m{m( m+n)" - (6m' +6mn+n") -1n+ 22+28}

The deficiency is g-ivenby-2p'=1n'+K'-2n'+2 ( 135)=m(m+n-8) +t-2m(m-I) +2=(m-I)(7n-2)+2 mpp'=H7n-l)(1n-2) +mp.

164. There is 110 difficulty in seeing how these numbersare to be modified, if the original curve touches the line atinfinity 01' passes through the circular points at infinity.

Thus, if the line at infinity touches the curve at gdifferent points, we have-

n'=(m-'.f)(m-I) m'=(-m-y)(m+n-:J-g)K'=('In-y)t; uud so OIl.

In order I,u obtaiu these results we recall Lhe results uf 143. It will be noticed that if the curve touches the lineat infinity, the absolute term will 1101, appear in itstaugeut.iul equation, aud the co-efficients ill the eliminantare of degree (m-I), and the degree of the orthopt.ic locusis, therefore, (1 n -1)".

Hence, the orthoptic locus of It circle is a circle, whileLhlL t of ;t parabola is :1 straight line (the directrix).

If, however, the lineal' as well as the absolute term areabsent, the line at infinity is It bitaugent 01' a stationarytangent, and the co-efficients ill the eliminant. are of degree(m-2), and the orthoptic locus is of order (m-2)(m-I).


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206 THBORY 01' PLANE CURVES

And geuerally, If the line at infinity i" a 1IL1~ltiple tangentIf order g, the degree of tilt> orthoptic lOC1I8 (~i it CItTVe ( ~ f thinth. class is (m-gHm-l).III It similar manner, tit e class is found to be-

m'= (?n-g)( 111 +n-:3 -g)The bugeutial equation of the evolute of the parabola

4ae =27T/2.~Here the liuear and the absolute terms are absent, an.l LIteorthoptic locus is a parabola.

The line at infinity is a bitangent to the evolute of theellipse, and hence the orthoptic locus is a sexbic curve.

Dr. C. 'I'ay 101' ill a note, published in the Proc, of the H,0 .Y a1Society of I..ondou, Vol. :37 (1884), pp. 138-141, propoundeda number of theorems on the isoptic and orthoptic loci of acurve, where he remarks that these had been verified bj'analytical methods ill an unpublished paper by one Mr. J. KYea. The theorems stated here are taken from the said note.

In a similar manner, the Plucker's numbers are to bemodified when the curve passes through I and J, or hasmultiple points at those points.

Ere. 1. Show that the orthoptic locus is a circle, when the curve i8/l . central conic (or a circle).

EJ!. 2. Show that the orthoptic locus is a straight, line, when thecurve is a parabola.

Ere. 3. The orthoptic. locus of a quartic curve of class 3, touching theliue at infinity at two points dividing IJ harmonically, is a straight line.

Ex. 4. Find the characteristics of the orthoptic locus when theinflexional tangents of the curve pass through the circular points.

[n'=(m+l)(m-2), m'=,n(m+n-4), ,,'="1I-4.JEx. 5. Find the characteristics of t.he orthoptic locus of the evolnte

ora curve .. [The locus is the same as the locus of intersections of perpendicular

normals of the curve. Hence,1/:=m-1(m+ -2), m'=(",-1)(4", +"-6), etc.}


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cHARACTERISTICS OF cuuvxs 207165. THE CUAHACTERISTICS OF Al\ !sOPTIG Locus s :Since each of the circular points is an m(m-l)-ple point

on the isopbic locus, proceeding in It manner similar to thatin 162, it is found that the degree of the isoptic IOCUR iR-

n'=2m(m-1'i,To find the class, the tangents at iare regarded as2m(m-1) tangents which can be drawn from I to the curve;and proceeding as in 168, it will be found that there areother 2m(n-1) tangents which can be drawn from 1 tothe curve.

Thus,m'=2m(m-1) +2m(n-1) =2m(m+n-2)Also, K'=2mtFrom these we ma.ycalculate the other characteristics,Thus, S'=m(2m-3)(m-m-1)+2mT.

T'=m{2m(m+n)' -(Sm' +Smn+n') -2m+ 12+2S}.t'=2m(3nt+K--3) p'=(m-l)' +2mp.

Ex. 1. Find the character-istics of the envelope of a chord of acurve which aubtends an angle of given magnitude at l\ given point.

Ex. 2. Prove that the envelope of a circle which passes through afixed point and subtends a constant angle at another is a limacon.

Ex. 3. The isoptic locus of a parabola consists of the line at infinitytwice and It central conic (Taylor ) .

Ex. 4. Prove that the points of contact of the tangents drawnfrom the circular points to any curve are Ringle points on the orthopticJOC U R, and double points on [sop tic locus (Tay/o r ) .

166. OTHEl~ DERIVEV CUlW}


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THEORY OF PLANE CllRV~;S

Properties of such cur-ves may be discussed indepenrientlyas occasion arises. \Ve shall, however, conclude this chapterwith a few examples, which t,he students are reqnil'flflto work out for themselves.

Ex. 1. If on the radius OP of a straig'ht line, we mensnre all'distaneea PQ= k , the loons of Q is " curve having' n norln at ()nnil n - tncnode at infinity with the given line as tangent.

[The locus is culled the Conchoid of Nicomedos.]E, 2. Find the locns of a point Q taken on tho radius veotm- OP

to a (li1'('I"through the point 0 snch that PQ = k,['I'his locus is called Pnscal's Limacon.]

E:r.:'I. TliSCIlASho natrn-e of tho origin in Ex. 2.E." . 4. Any stra.ight linc OP intersects two given ClUVOS :\.1. 1'1 llnd

P,. Find the locus of a point P snch thnt OP=OP,-OP,.['rhis curve is called the Cissoid of tho curves for th" polr- O. J

E.". G. II'im1tho ehru-actm-isnios of the locus in Jilx. 4.[n'=2n1'n2,

where nuJnu "1' etc., and n" m2, "2' etc., are tho chn.ractol ' iRtiC8: orthe two given curves.]

Em. 6. Find the locus of P and its charncteristics, if in Ex. G,OP=k,OP, +k.OP., where ""k2 are constants.Ex. 7. Show that the cissoid of two circles, one of which pnssos

through the origin, is a quartic curve having a node at the origin andtwo nodes at the circular points (A bi-circnlar quarnic ).

E, 8. If on the radius vector at' of a curve, PQ= " is measuredoff, where k is a constaut, the locus of Q is the cissoid of the givf>ncurve and the circle with centre ani! radius k;

[This curve is called the conchoid of the curve. Of, Ex. 1.]E, 9. Find the oonohoid for the conic given by tho genol'ai

equation of the second degree.Ex. 10. Show that tho characteristics of the locus of the centre

of a circle orthogonal to a gi,'en circle and touching a given 2".i,with n.ple points at I and J are the same as those of the reciprocalto the given 2n."c.

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Ganguli, Theory of Plane Curves, Chapter 7

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